## Sunday, 4 January 2015

### 3.4 Number of Moles and Its Molar Volume

Number of Moles and Its Molar Volume
Definition : : Molar volume : The volume occupied by 1 mole of gas at a particular temperature and pressure.
 ► At room conditions (25°C and pressure 1 atmosphere), 1 mole of gas occupies a volume of 24dm3 . ► At standard condition (0 and pressure 1 atmosphere), 1 mole of gas occupies a volume 22.4dm3.
■ Summary of relationship between the number of moles, the volume of gas and the molar volume of gas.

Worked-example 3.4(a)
 Calculate the volume occupied by 3.5 moles of ammonia gas, ${\text{NH}}_{3}$ at s.t.p. Solution Volume of gas $= 3.5×22.4$ = 78.4dm3

Worked-example 3.4(b)
 Calculate the volume occupied by the following gases at s.t.p: [RAM: H = 1, C = 12, N = 14, O = 16,], [NA = $6.02×{10}^{23}$ , 1 mole gas occupies 22.4dm³ at s.t.p] (a) 8.8 g carbon dioxide, Solution Step 1: $\begin{array}{c}\text{Number of moles =}\frac{\text{Mass of Carbon Dioxide}}{\text{Relative molecular mass}}\end{array}$ $\begin{array}{c}\text{=}\frac{8.8}{12 + \left(2×16\right)}\end{array}$ $\begin{array}{c}\text{=}\frac{8.8}{12 + 32}\end{array}$ $\begin{array}{c}\text{=}\frac{8.8}{44}\end{array}$ = 0.2mol Step 2: Volume of 8.8 g carbon dioxide, ${\text{CO}}_{2}$ $\text{= Number of mole}×\text{Molar volume}$ $= 0.2×22.4$ = 4.48dm3 (b) $1.5×{10}^{23}$ molecules of oxygen, Solution Step 1: $\begin{array}{c}\text{Number of moles =}\frac{\text{Number of particles}}{\text{Avogrado constant}}\end{array}$ $\begin{array}{c}\text{=}\frac{1.5×{10}^{23}}{6×{10}^{23}}\end{array}$ = 0.25mol Step 2 Volume of oxygen $\text{= Number of mole}×\text{Molar volume}$ $= 0.25×22.4$ = 5.6dm3
Summary of Mole Concept
■ Summary of relationship between the number of moles of compound with the number of particles, particle mass and gas volume.

Worked-example 3.4(c)
 Calculate the volume occupied by the following gases at s.t.p. [Molar volume at s.t.p = 22.4dm3, NA = $6×{10}^{23}$ , H = 1, C = 12, O = 16, N = 14, Cl = 35.5, Ne = 20] (a) 0.5 g of neon , Ne. Solution Step 1: $\begin{array}{c}\text{Number of moles =}\frac{\text{Mass of neon}}{\text{Relative atom mass}}\end{array}$ $\begin{array}{c}\text{=}\frac{0.5}{20}\end{array}$ = 0.025mol Step 2: Volume of 0.5 g neon ${\text{CO}}_{2}$ $\text{= Number of mole}×\text{Molar volume}$ $= 0.025×22.4$ = 0.56dm3 (b) 1.5 moles of carbon dioxide, ${\text{CO}}_{2}$ Solution : Volume of 1.5 moles of ${\text{CO}}_{2}$ $\text{= Number of mole}×\text{Molar volume}$ $= 1.5×22.4$ = 33.6dm3 (c) $4.8×{10}^{22}$ molecules Chloride, ${\text{Cl}}_{2}$ Solution Step 1: $\begin{array}{c}\text{Number of moles =}\frac{\text{Number of particles}}{\text{Avogrado constant}}\end{array}$ $\begin{array}{c}\text{=}\frac{4.8×{10}^{22}}{6×{10}^{23}}\end{array}$ = 0.08mol Step 2: Volume of ${\text{Cl}}_{2}$ $\text{= Number of mole}×\text{Molar volume}$ $= 0.08×22.4$ = 1.792dm3